In the diagram,I_1 and I_2 are the strengths | vedclass

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(D) The magnetic field at the centre $O$ due to the circular loop of radius $R$ carrying current $I_1$ is given by $B_1 = \frac{\mu_0 I_1}{2R}$ (direc

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$\frac{1}{2\pi}$

Vedclass · Answer

(D) The magnetic field at the centre $O$ due to the circular loop of radius $R$ carrying current $I_1$ is given by $B_1 = \frac{\mu_0 I_1}{2R}$ (directed outwards).The magnetic field at point $O$ due to the infinitely long straight wire carrying current $I_2$ at a distance $d = OA + AB = R + R = 2R$ is given by $B_2 = \frac{\mu_0 I_2}{2\pi d} = \frac{\mu_0 I_2}{2\pi (2R)} = \frac{\mu_0 I_2}{4\pi R}$ (directed inwards).Since the net magnetic field at $O$ is zero,the magnitudes of the magnetic fields must be equal:$B_1 = B_2$$\frac{\mu_0 I_1}{2R} = \frac{\mu_0 I_2}{4\pi R}$Canceling common terms $\mu_0$ and $R$ from both sides:$\frac{I_1}{2} = \frac{I_2}{4\pi}$Rearranging to find the ratio $I_1/I_2$:$\frac{I_1}{I_2} = \frac{2}{4\pi} = \frac{1}{2\pi}$Therefore,the correct option is $D$.

In the diagram,$I_1$ and $I_2$ are the strengths of the currents in the loop and the infinitely long straight conductor,respectively. Given $OA = AB = R$. The net magnetic field at the centre $O$ is zero. Then the ratio of the currents in the loop and the straight conductor $(I_1/I_2)$ is:

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